What is the smallest cardinality a topology can have which is c.c.c but not separable (in ZFC)?

This may be helpful for you:
Proof: $X$ is not separable: for any countable set $A \subset X$, clearly, $U=X\setminus A$ is open and $U \cap A=\emptyset$. $X$ is CCC: if $X$ has uncountable disjoint open sets $\lbrace \cal U_\xi: \xi \in \aleph_1\rbrace$. Pick one open set, for example, $U_0$. Because $X \setminus U_0$ is uncountable, it is a contradiction with $U_0$ is open. 


If by "cardinality of the topology" you mean "cardinality of the underlying space" then the answer is $\aleph_1$. Let $2^{\omega_1}$ be the product of $\omega_1$ many copies of the 2point discrete space and consider the subset $X \subset 2^{\omega_1}$ consisting of all points with at most finitely many nonzero entries. Since $X$ is dense in the separable space $2^{\omega_1}$, $X$ is ccc, but it's easy to see that $X$ is not separable. Now $X$ has cardinality $\aleph_1$. If by "cardinality of the topology" you mean "cardinality of the set of all open sets" then it depends. If you don't care about your spaces being Hausdorff, then the answer is again $\aleph_1$: just consider the topology on $\omega_1$ consisting of all final segments and the empty set. If you want your spaces to be Hausdorff, note that every infinite Hausdorff space has at least continuum many distinct open sets (just take an infinite countable pairwise disjoint family of nonempty open sets and consider all possible unions). But actually, there is a ccc nonseparable space with exactly continuum many open sets in ZFC. This is Justin Moore's Lspace (an example of a regular hereditarily Lindelof nonseparable space). This space is ccc because it is hereditarily Lindelof and it is a subspace of the product of $\omega_1$ many copies of the circle, and hence it has a base of cardinality $\aleph_1$. Since it's hereditarily Lindelof, every open set is the union of countably many open sets from that base. Therefore it has at most continuum many open sets. So the minimum number of open sets of a Hausdorff ccc nonseparable space is continuum. 


The forcing $\text{Add}(\omega,\aleph_1)$ to add $\aleph_1$ many Cohen reals is c.c.c. but not separable, and has size $\aleph_1$, when it is considered as a poset rather than as a complete Boolean algebra. This poset is the set of finite partial functions from $\omega_1$ to $2$, ordered by extension. But every countable space is of course separable, so the answer is $\aleph_1$. 

